Divide the following complex numbers. $ \dfrac{4-2i}{2+4i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${2-4i}$ $ \dfrac{4-2i}{2+4i} = \dfrac{4-2i}{2+4i} \cdot \dfrac{{2-4i}}{{2-4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(4-2i) \cdot (2-4i)} {(2+4i) \cdot (2-4i)} = \dfrac{(4-2i) \cdot (2-4i)} {2^2 - (4i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(4-2i) \cdot (2-4i)} {(2)^2 - (4i)^2} = $ $ \dfrac{(4-2i) \cdot (2-4i)} {4 + 16} = $ $ \dfrac{(4-2i) \cdot (2-4i)} {20} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({4-2i}) \cdot ({2-4i})} {20} = $ $ \dfrac{{4} \cdot {2} + {-2} \cdot {2 i} + {4} \cdot {-4 i} + {-2} \cdot {-4 i^2}} {20} $ Evaluate each product of two numbers. $ \dfrac{8 - 4i - 16i + 8 i^2} {20} $ Finally, simplify the fraction. $ \dfrac{8 - 4i - 16i - 8} {20} = \dfrac{0 - 20i} {20} = -i $